Sciencemadness Discussion Board

Зaйцев's rule

pilgrim - 16-11-2009 at 15:57

The reactions of haloalkanes form the major product (i.e. internal alkene) beacuse it is more stable. Hoffman's products are formed when a base has difficulty absracting a proton, either because the base itself is sterically hindered or because the proton whose abstraction would lead to the more stable alkene is hindered from approach. (Confirm?)

The question I need help figuring out is as follows.
Supposed we have a unimolecular elimination of 2-bromo butane. Is it the nature of the chemicals to form the most stable products when the products are of similar structure, or do reactions depend instead on the probabilty in this way: the carbocation can form the double bond through either of two electron pairs from an adjacent carbon atom, that is not the terminal carbon atom which has 3 pairs of electrons that could be used. Why do we not expect initial products in a 3:2 ratio (solely based on the probability of hydrogen abstraction), then reason that the Zaitsev product forms after the system reaches equilibrium.

If we run this particular experiment, does the molecule form a terminal alkene, then after equilibrium, turn into the more stable product? Or, is it already natural for the chemicals to form the bonds that would already lead to the most stable aklene.

With someone with pracitcal knowledge tell me how quickly this experiment would yeild the major product? I think I am trying to emphasize someone give me an argument against the transittion state formation being dependant on the probability of abstraction.

I hope you can understand that i don't do this often, and am sorry if and of this is unclear. I'm obviously having difficulty getting an understanding and am looking forward to an understandable answer. Thank you.

sonogashira - 16-11-2009 at 17:08

I believe the kinetic product is first formed, then if there is enough energy (temperature) it can convert into the thermodynamic product [but if there is enough energy to begin, the higher-energy initial reaction can occur]. This is why very cold temperatures are often used to abstract a less-hindered proton (which would rearrange to the more stable thermodynamic product at room temperature). If there is not the activation energy to rearrange to the thermodynamic product, it will persist.

In your specific example I think there is likely very little (practical) difference in the acidity (and hinderence) of the two kinds of proton.
The mechanism will also depend on the solvent, and the base... and the interaction between the two at the temperature and pressure of the laboratory. In reality there will probably be more than one mechanism occurring, even if only a few percent... and rearrangements also... so I feel that one can only use the theory as a guide. And as you say there is also the probability, and the acidity of each type of proton, to consider.

I don't know if this helps and maybe I have confused you even more... as I have myself! So sorry, hehe! :)

[Edited on 17-11-2009 by sonogashira]

jon - 16-11-2009 at 20:45

this is thermodymnamic rxn control. you can conrol rates w/ temp. and concentration. kinetic control.

pilgrim - 17-11-2009 at 05:42

Being a thermodynamic reaction control, this is to say, the more stable product will form right away?

Nicodem - 17-11-2009 at 12:31

Quote: Originally posted by pilgrim  
The reactions of haloalkanes form the major product (i.e. internal alkene) beacuse it is more stable. Hoffman's products are formed when a base has difficulty absracting a proton, either because the base itself is sterically hindered or because the proton whose abstraction would lead to the more stable alkene is hindered from approach. (Confirm?)

Both, either the E1 or E2 elimination of alkyl (pseudo)halides give mainly that alkene which is thermodynamically more stable (is at lower energy). The Hoffman elimination mechanism is different and makes the alkene resulting from deprotonation at the most acidic beta-hydrogen. It therefore generally gives an opposite regioselectivity. Steric hindrance of the base has no role in the Hoffman elimination, while in the E2 elimination stericaly hindered bases are used only to prevent the SN2 side reaction to prevail. In E1 eliminations the bulkiness of the base is not important.
Quote:
The question I need help figuring out is as follows.
Supposed we have a unimolecular elimination of 2-bromo butane. Is it the nature of the chemicals to form the most stable products when the products are of similar structure, or do reactions depend instead on the probabilty in this way: the carbocation can form the double bond through either of two electron pairs from an adjacent carbon atom, that is not the terminal carbon atom which has 3 pairs of electrons that could be used. Why do we not expect initial products in a 3:2 ratio (solely based on the probability of hydrogen abstraction), then reason that the Zaitsev product forms after the system reaches equilibrium.

You must consider that in E1 elimination (unimolecular elimination) the carbocation eliminates a proton in a concerted manner, meaning that the C-H bonds breaks synchronously as the C=C double bond is forming. Therefore the elimination reaction of the proton which leads to the lowest energy alkene will also have the lowest activation energy (in this mechanism the Ea is related to the sum of all energies of broken and created bonds). For example, lets say the case is of E1 elimination of 2-bromo-2-methylbutane. Once the carbocation forms, either the H on one of the two geminal methyls can eliminate, lets call it H(a), or the H(b) from the methylene group. Both C-H(a) and C-H(b) bond energies are very similar, but the bond energy of the new C=C bond forming at the two possible alkenes are quite different. Consider the elimination of the H(a) proton: this forms a terminal alkene having two geminal alkyls. The elimination of H(b) gives an alkene having three methyls in a 1,1,2-substitution pattern. Since the thermodinamic stability (energy level) of an alkene is lower the more electron donating alkyl substituents it has, the C=C bond formation in the case of H(b) elimination is more energetic, thus lowering the activation energy due to synchronicity.
In E2 elimination the situation about the activation energy is practically identical, because the C-Br bond energy is a constant for both pathways and again mainly the C=C bond energy dictates the regioselectivity.
Quote:
If we run this particular experiment, does the molecule form a terminal alkene, then after equilibrium, turn into the more stable product? Or, is it already natural for the chemicals to form the bonds that would already lead to the most stable aklene.

The regioselectivity is inherent to the mechanism.
This has nothing to do with kinetic vs. thermodinamic control. You can only talk about kinetic and thermodinamic products when you have a reaction of the type A => B => C where the A => B is faster than B => C, which is not the case in E1 and E2 eliminations where the alkenes do not rearrange once they form. At least usually not - the double bond in some alkenes can rearrange in basic enough conditions in certain substrates with acidic enough allylic position. In such a case you could have a A => B => C transformation, but this is not the case in simple olefins.
Quote:
With someone with pracitcal knowledge tell me how quickly this experiment would yeild the major product? I think I am trying to emphasize someone give me an argument against the transittion state formation being dependant on the probability of abstraction.

The term abstraction is almost exclusively used for homolytic bond cleavage (like in radical abstractions of a hydrogen, for example), so I would avoid using it here.
E1 eliminations on tert-alkyl halides are very rapid. They occur already at room temperature in polar enough solvents and with moderately strong bases. The E2 elimination requires heating and works best in aprotic solvents like DMF, THF and so on, where bases are stronger (and where even non-bulky bases give relatively good results). For example, if you heat a primary alkyl halide and NaOMe in DMF you will most commonly get the alkene as the major product, but if you heat it in MeOH as solvent, then you get the methyl ether resulting from the SN2 substitution as the major product. With secondary alkyl halides the competition with the SN2 substitution is much less of a problem.
Quote:
I hope you can understand that i don't do this often, and am sorry if and of this is unclear. I'm obviously having difficulty getting an understanding and am looking forward to an understandable answer. Thank you.

Don't worry, it is normal that you want an answer for something that is not, or is usually not in the textbooks. We teach students a lot of terribly simplified theories, because students are usually not interested to understand anything. They just need something to memorize in order to pass the exam. But when you work in a lab it is quite different from being a student interested only in grades. Then you are expected to learn theory by working and reading, because when faced with the real world situations you soon learn that without theory you can not do anything good.

kmno4 - 17-11-2009 at 17:10

Exemplary literature to read:
Angewandte Chemie International Edition, Volume 48 Issue 31, Pages 5724 - 5728.

Also:
Solvents and Solvent Effects in Organic Chemistry, 3rd, Updated and Enlarged Edition (Wiley)
And:
Russian Chemical Reviews 53, 547 (1984)
... and many other.